Entangled Two-Particle States

Section 8.2 Entangled Two-Particle States

For a single electron, any possible spin state can be written as

\begin{matrix} (8.1) & | ψ ⟩ = c_{+} | + z ⟩ + c_{-} | - z ⟩ = c_{+} | ↑ ⟩ + c_{-} | ↓ ⟩, \end{matrix}

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with the appropriate choice of the complex numbers

c_{+}

and

c_{-} .

For notational simplicity we will switch to writing the spin states in terms of up and down arrows for the remainder of this chapter.

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Now consider a two-particle state made up of two different kinds of particle. To be specific, let's take an electron and a positron, which is a particle with the same mass and spin as the electron, but it has a

+ e

charge.¹ What is the expression analogous to (8.1) for any possible two-particle state?

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Table 8.1. The four possible states with definite

S_{z}

values


state	electron	positron

$\| ↑↑ ⟩ = \| ↑ ⟩ \| ↑ ⟩$	$S_{z} = + ℏ / 2$	$S_{z} = + ℏ / 2$
$\| ↑↓ ⟩ = \| ↑ ⟩ \| ↓ ⟩$	$S_{z} = + ℏ / 2$	$S_{z} = - ℏ / 2$
$\| ↓↑ ⟩ = \| ↓ ⟩ \| ↑ ⟩$	$S_{z} = - ℏ / 2$	$S_{z} = + ℏ / 2$
$\| ↓↓ ⟩ = \| ↓ ⟩ \| ↓ ⟩$	$S_{z} = - ℏ / 2$	$S_{z} = - ℏ / 2$

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There are four possible states that have definite values for the

z

-component of the spin, listed in Table 8.1. These four states make a basis in the sense that any possible two-particle state for the electron and positron is a superposition of these states:

\begin{matrix} (8.2) & | ψ ⟩ = c_{1} | ↑↑ ⟩ + c_{2} | ↑↓ ⟩ + c_{3} | ↓↑ ⟩ + c_{4} | ↓↓ ⟩ . \end{matrix}

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If the two particles were both electrons, we would need to build in indistinguishability by choosing coefficients

c_{1}

c_{4}

that make an anti-symmetric state. But that's not the case here: particle 1 is an electron and particle 2 is a positron, so they are distinguishable from each other. The state vector does not need to be anti-symmetric.

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We have also introduced in Table 8.1 the notation that we can write a two-particle state with definite values of

S_{z}

in a factored form:

| ↑↓ ⟩ = | ↑ ⟩ | ↓ ⟩ .

This is simply two different ways to write the same thing: the electron has spin up and the positron has spin down. Both notations have their advantage and we'll switch back and forth between them. But be careful when using the “factored” states: the electron state vector must always be on the left and the positron state vector on the right. In other words,

| ↑ ⟩ | ↓ ⟩ \neq | ↓ ⟩ | ↑ ⟩ .

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Now let's examine the behavior of this two particle state with an example. Consider an electron-positron pair in the state

\begin{matrix} (8.3) & | ψ ⟩ = 0.9 | ↑↑ ⟩ + 0.1 | ↑↓ ⟩ + 0.3 | ↓↑ ⟩ + 0.3 | ↓↓ ⟩ . \end{matrix}

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First, let's compute the probabilities associated with a measurement of the positron's

S_{z}

spin component.

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There are two ways that the positron could be found to have

S_{z} = + ℏ / 2,

as both the

| ↑↑ ⟩

state and the

| ↓↑ ⟩

state have the positron spin up. So we need to add together the probability of measuring

| ↑↑ ⟩

and the probability of measuring

| ↓↑ ⟩ :

\begin{aligned} Prob(positron spin up) & = Prob (| ↑↑ ⟩) + Prob (| ↓↑ ⟩) \\ = | 0.9 |^{2} + | 0.3 |^{2} \\ = 0.81 + 0.09 = 0.90 . \end{aligned}

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Note that we did not add the coefficients first and then square them. That would have led to the

| 0.9 + 0.3 |^{2} = 1.44,

which is a nonsensical result for a probability!

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Similarly, we calculate the probability of the positron being found with spin down by combining the probabilities of the

| ↑↓ ⟩

and

| ↓↓ ⟩

states:

\begin{matrix} (8.4) & Prob(positron spin down) = | 0.3 |^{2} + | 0.1 |^{2} = 0.10 . \end{matrix}

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Notice that we have a normalized state and the total probability of the positron having either spin up or spin down adds to one:

0.90 + 0.10 = 1 .

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Now let's pose a question that gets at the central topic of this chap:

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Suppose we start out again with the state $| ψ ⟩$ and first measure the electron's $S_{z}$ value. Will this measurement affect the state of the positron?

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Here's a way to approach it. Let's write the basis states in their factored form, and see if we can combine terms:

\begin{aligned} | ψ ⟩ & = 0.9 | ↑ ⟩ | ↑ ⟩ + 0.1 | ↑ ⟩ | ↓ ⟩ + 0.3 | ↓ ⟩ | ↑ ⟩ + 0.3 | ↓ ⟩ | ↓ ⟩ \\ = | ↑ ⟩ (0.9 | ↑ ⟩ + 0.1 | ↓ ⟩) + | ↓ ⟩ (0.3 | ↑ ⟩ + 0.3 | ↓ ⟩) . \end{aligned}

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Look at what we have done here: we have used a distributive rule to factor our two-particle state. This is allowed, but keep in mind that in any product we must keep the electron state vector on the left and the positron state vector on the right.

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To make this expression more useful, we would like the terms in parentheses to be normalized states. We can achieve this by multiplying and dividing by the appropriate factor, something like

\begin{matrix} (8.5) & | ψ ⟩ = c_{+} | ↑ ⟩ (\frac{0.9}{c_{+}} | ↑ ⟩ + \frac{0.1}{c_{+}} | ↓ ⟩) + c_{-} | ↓ ⟩ (\frac{0.3}{c_{-}} | ↑ ⟩ + \frac{0.3}{c_{-}} | ↓ ⟩) . \end{matrix}

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As you will check in the homework, the terms in parentheses will be normalized states as long as

c_{+} = \sqrt{0.82}

and

c_{-} = \sqrt{0.18} .

Putting in those values gives

\begin{matrix} (8.6) & | ψ ⟩ = \sqrt{0.82} | ↑ ⟩ | ϕ_{1} ⟩ + \sqrt{0.18} | ↓ ⟩ | ϕ_{2} ⟩ \end{matrix}

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with the normalized positron states

\begin{aligned} | ϕ_{1} ⟩ & = \frac{9}{\sqrt{82}} | ↑ ⟩ + \frac{1}{\sqrt{82}} | ↓ ⟩ \\ (8.7) & | ϕ_{2} ⟩ & = \frac{1}{\sqrt{2}} | ↑ ⟩ + \frac{1}{\sqrt{2}} | ↓ ⟩ . \end{aligned}

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Now,

| ψ ⟩

written in the form (8.6) is nothing other than the

| ψ ⟩

we started with in (8.3). But it is now much more useful: we can finally address our central question (does an electron spin measurement affect the positron state) using tools we've already learned for states.

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An electron

S_{z}

measurement will have a probability of

0.82

of finding the electron spin up. If this occurs, the state will collapse to the new state

\begin{matrix} (8.8) & | ψ_{new} ⟩ = | ↑ ⟩ | ϕ_{1} ⟩ . \end{matrix}

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Alternately, there is a probability

0.18

of the electron being found to have spin down, in which case the state collapses to

\begin{matrix} (8.9) & | ψ_{new} ⟩ = | ↓ ⟩ | ϕ_{2} ⟩ . \end{matrix}

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We can now determine the probabilities for the subsequent positron spin measurement by looking at the coefficients in the normalized states

| ϕ_{1} ⟩

and

| ϕ_{2} ⟩ .

And here is what we find:

If we do not measure the electron's spin, the positron has a probability $9 / 10$ of being spin up.
If we measure the electron's spin and find it to be spin up, then the positron has a probability $81 / 82$ of being spin up.
If our electron spin measurement had instead found it to be spin down, then the positron has a probability $1 / 2$ of being spin up.

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This is the essence of entanglement! Measurements on one particle affect the state of the other particle.

You may have heard of positrons at least indirectly: they are used in Positron Emission Tomography, also known as a PET scan.